Question

evaluate the following sum.
\\(\sum\limits_{k=1}^{3} \frac{k^3}{9} + \left(\sum\limits_{k=1}^{3} k\
ight)^2\\)

\\(\sum\limits_{k=1}^{3} \frac{k^3}{9} + \left(\sum\limits_{k=1}^{3} k\
ight)^2 = 40\\)
(simplify your answer.)

Explanation:

Step1: Evaluate the first sum $\sum_{k = 1}^{3}\frac{k^{3}}{9}$

For $k = 1$: $\frac{1^{3}}{9}=\frac{1}{9}$
For $k = 2$: $\frac{2^{3}}{9}=\frac{8}{9}$
For $k = 3$: $\frac{3^{3}}{9}=\frac{27}{9} = 3$
Sum these values: $\frac{1}{9}+\frac{8}{9}+3=\frac{1 + 8}{9}+3=1 + 3=4$

Step2: Evaluate the second sum $\sum_{k = 1}^{3}k$

For $k = 1$: $1$
For $k = 2$: $2$
For $k = 3$: $3$
Sum these values: $1+2 + 3=6$

Step3: Square the second sum and add to the first sum

Square the result of the second sum: $6^{2}=36$
Add to the first sum: $4+36 = 40$

Answer:

40